/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 The resistance of a conductor is... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 18: Problem 28

The resistance of a conductor is \(19.8 \Omega\) at \(15.0^{\circ} \mathrm{C}\) and \(25.0 \Omega\) at \(85.0^{\circ} \mathrm{C} .\) What is the temperature coefficient of resistance of the material?

Short Answer

Expert verified
Answer: The temperature coefficient of resistance of the material is approximately \(3.72 \times 10^{-3} \ \mathrm{K}^{-1}\).

Step by step solution

01

Identify the given values

We are given the resistances of the conductor at two temperatures: - \(R_1 = 19.8 \Omega\) at \(T_1 = 15.0^{\circ} \mathrm{C}\) - \(R_2 = 25.0 \Omega\) at \(T_2 = 85.0^{\circ} \mathrm{C}\)
02

Convert the temperatures to Kelvin scale

To calculate the temperature coefficient of resistance, it is essential to use the Kelvin scale for temperature. The conversion from Celsius to Kelvin is as follows: \(K = ^{\circ}\mathrm{C} + 273.15\). So, the temperatures in Kelvin are: - \(T_1 = 15.0^{\circ} \mathrm{C} + 273.15 = 288.15 \mathrm{K}\) - \(T_2 = 85.0^{\circ} \mathrm{C} + 273.15 = 358.15 \mathrm{K}\)
03

Substitute the given values into the temperature coefficient of resistance formula

Now that we have the resistances and corresponding temperatures in Kelvin, we can calculate the temperature coefficient of resistance (\(\alpha\)) using the following formula: \(\alpha = \dfrac{(R_2 - R_1)}{R_1(T_2 - T_1)}\) Substitute the given values into the formula as follows: \(\alpha = \dfrac{(25.0 \Omega - 19.8 \Omega)}{19.8 \Omega(358.15 \mathrm{K} - 288.15 \mathrm{K})}\)
04

Simplify and solve for the temperature coefficient of resistance

First, simplify the equation: \(\alpha = \dfrac{5.2 \Omega}{19.8 \Omega(70 \mathrm{K})}\) Next, cancel out the units and solve for \(\alpha\): \(\alpha = \dfrac{5.2}{19.8 \times 70}\) \(\alpha \approx 3.72 \times 10^{-3} \ \mathrm{K}^{-1}\)
05

Write down the final answer

The temperature coefficient of resistance of the material is approximately \(3.72 \times 10^{-3} \ \mathrm{K}^{-1}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An ammeter with a full scale deflection for \(I=10.0 \mathrm{A}\) has an internal resistance of \(24 \Omega .\) We need to use this ammeter to measure currents up to 12.0 A. The lab instructor advises that we get a resistor and use it to protect the ammeter. (a) What size resistor do we need and how should it be connected to the ammeter, in series or in parallel? (b) How do we interpret the ammeter readings?
A wire with cross-sectional area \(A\) carries a current \(I\) Show that the electric field strength \(E\) in the wire is proportional to the current per unit area \((I / A)\) and identify the constant of proportionality. [Hint: Assume a length \(L\) of wire. How is the potential difference across the wire related to the electric field in the wire? (Which is uniform?) Use \(V=I R\) and the connection between resistance and resistivity.]
If a chandelier has a label stating \(120 \mathrm{V}, 5.0 \mathrm{A},\) can its power rating be determined? If so, what is it?
A portable CD player does not have a power rating listed, but it has a label stating that it draws a maximum current of \(250.0 \mathrm{mA}\). The player uses three \(1.50-\mathrm{V}\) batteries connected in series. What is the maximum power consumed?
About \(5.0 \times 10^{4} \mathrm{m}\) above Earth's surface, the atmosphere is sufficiently ionized that it behaves as a conductor. The Earth and the ionosphere form a giant spherical capacitor, with the lower atmosphere acting as a leaky dielectric. (a) Find the capacitance \(C\) of the Earth-ionosphere system by treating it as a parallel plate capacitor. Why is it OK to do that? [Hint: Com- pare Earth's radius to the distance between the "plates." (b) The fair-weather electric field is about \(150 \mathrm{V} / \mathrm{m},\) downward. How much energy is stored in this capacitor? (c) Due to radioactivity and cosmic rays, some air molecules are ionized even in fair weather. The resistivity of air is roughly \(3.0 \times 10^{14} \Omega \cdot \mathrm{m}\) Find the resistance of the lower atmosphere and the total current that flows between Earth's surface and the ionosphere. [Hint: since we treat the system as a parallel plate capacitor, treat the atmosphere as a dielectric of uniform thickness between the plates.] (d) If there were no lightning, the capacitor would discharge. In this model, how much time would elapse before Earth's charge were reduced to \(1 \%\) of its normal value? (Thunderstorms are the sources of emf that maintain the charge on this leaky capacitor.)
See all solutions

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Read Explanation

Atoms and Radioactivity

Read Explanation

Electricity

Read Explanation

Waves Physics

Read Explanation

Physics of Motion

Read Explanation

Space Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.